Describe how you would prepare each of the following aqueous solutions: \((\mathbf{a}) 1.50 \mathrm{~L}\) of $0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ solution, starting with solid $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} ;(\mathbf{b}) 225 \mathrm{~g}\( of a solution that is \)0.65 \mathrm{~m}\( in \)\mathrm{Na}_{2} \mathrm{CO}_{3},\( starting with the solid solute; \)(\mathbf{c}) 1.20$ \(\mathrm{L}\) of a solution that is $15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}$ by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 \mathrm{M}\) solution of \(\mathrm{HCl}\) that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with $6.0 \mathrm{MHCl}$.

Calculate the quantity (moles) of solute required and further convert it into grams using the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). Molarity (M) = moles of solute / volume of solution in liter Rearranging the formula, moles of solute = Molarity × volume of solution in liter moles of solute = 0.110 M × 1.50 L = 0.165 moles of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) Now, the molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) = \((2\times18) + (4\times14) + 4\times16 + 32\) = 132 g/mol So, the mass of solute (in grams) = moles of solute × molar mass = 0.165 moles × 132 g/mol = 21.78 grams To prepare 1.50 L of 0.110 M \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\), start by adding 21.78 grams of solid solute to a volumetric flask and then fill it up to the 1.50 L mark with water.

Convert the molality (m) into moles using the mass of the solution and then convert it into grams using the molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). Molality (m) = moles of solute / mass of solvent in kg Solving for moles of solute, moles of solute = molality × mass of solvent in kg Let x be the mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solute. Then the mass of the solvent (water) is (225 - x) grams. Convert this to kg by dividing by 1000: moles of solute = 0.65 m × (225 - x) / 1000 = 0.65 (225 - x) / 1000 Now, the molar mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) = \((2\times23) + 12 + (3\times16)\) = 106 g/mol Mass of solute = moles of solute × molar mass Thus, x = 0.65 (225 - x) / 1000 × 106 Solving for x, we get: x ≈ 11.21 grams To prepare 225 g of a 0.65 m \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution, start by adding approximately 11.21 grams of solid solute to a beaker and slowly adding water until the total mass of the solution is 225 g.

Calculate the mass of solute required using the mass percentage of the solution and then convert it into grams using the molar mass of \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\). Mass percentage of \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) = (mass of solute / mass of the solution) × 100 Mass of the solution = volume × density = 1.20 L × 1000 mL/L × 1.16 g/mL = 1392 g Now, let x be the mass of \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) solute. 15.0 = (x / 1392) × 100 Solving for x, we get: x ≈ 208.8 grams To prepare 1.20 L of a solution that is 15.0% \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) by mass, start by adding approximately 208.8 grams of solid solute to a volumetric flask and then fill it up to the 1.20 L mark with water.

We will first calculate the moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(\mathrm{HCl}\) needed for neutralization and then calculate the volume of 6.0 M \(\mathrm{HCl}\) needed. Finally, we will find the amount of water needed to dilute the \(\mathrm{HCl}\) solution to 0.50 M. The balanced chemical equation for the reaction between \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) is: $$\mathrm{Ba}(\mathrm{OH})_{2} + 2\mathrm{HCl} \rightarrow \mathrm{BaCl}_{2} + 2\mathrm{H}_{2}\mathrm{O}$$ Mole calculation: Moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\)/molar mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\) Moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = 5.5 g / (137.3 + 2 × (16 + 1)) g/mol ≈ 0.0205 mol From the stoichiometry, 1 mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) needs 2 moles of \(\mathrm{HCl}\) for neutralization: Moles of \(\mathrm{HCl}\) required = 2 × moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) = 2 × 0.0205 ≈ 0.041 mol Calculate the volume of 6.0 M \(\mathrm{HCl}\) needed: Volume of \(\mathrm{HCl}\) = moles of \(\mathrm{HCl}\) / Molarity ⇒ 0.041 mol ÷ 6.0 M ≈ 0.00683 L We want a 0.50 M \(\mathrm{HCl}\) solution. So let's find the final volume required: Final volume = moles of \(\mathrm{HCl}\) ÷ 0.50 M = 0.041 mol ÷ 0.50 M ≈ 0.082 L Now, find the amount of water needed: Amount of water needed = final volume - initial volume of \(\mathrm{HCl}\) = 0.082 L - 0.00683 L ≈ 0.0752 L To prepare a 0.50 M solution of \(\mathrm{HCl}\) that would just neutralize 5.5 g of \(\mathrm{Ba}(\mathrm{OH})_{2}\), start by taking 0.00683 L of 6.0 M \(\mathrm{HCl}\) and then add approximately 0.0752 L of water to dilute the \(\mathrm{HCl}\) solution to 0.50 M.